/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Do Babies Understand Probability... [FREE SOLUTION] | 91Ó°ÊÓ

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Do Babies Understand Probability? Can babies reason probabilistically? A study \(^{19}\) investigates this by showing ten- to twelve-month-old infants two jars of lollipop-shaped objects colored pink or black. Each infant first crawled or walked to whichever color they wanted, determining their "preferred" color. They were then given the choice between two jars that had the same number of preferred objects, but that differed in their probability of getting the preferred color; each jar had 12 in the preferred color and either 4 or 36 in the other color. Babies choosing randomly or based on the absolute number of their preferred color would choose equally between the two jars, while babies understanding probability would more often choose the jar with the higher proportion of their preferred color. Of the 24 infants studied, 18 chose the jar with the higher proportion of their preferred color. Are infants more likely to choose the jar with the higher proportion of their preferred color? (a) State the null and alternative hypotheses. (b) Give the relevant sample statistic, using correct notation. (c) Which of the following should be used to calculate a p-value for this dataset? A randomization test, a test using the normal distribution, or either one? Why? (d) Find a p-value using a method appropriate for this data situation. (e) Make a conclusion in context, using \(\alpha=0.05\).

Short Answer

Expert verified
Based on the Z-test and a p-value of 0.0072, which is less than the significance level \( \alpha \) of 0.05, we reject the null hypothesis. Therefore, it can be concluded that infants are more likely to choose the jar with a higher proportion of their preferred color.

Step by step solution

01

State the hypotheses

The null hypothesis (\(H_0\)) is infants are not more likely to choose the jar with the higher proportion of their preferred color (or infants choose randomly). The alternative hypothesis (\(H_A\)) is infants are more likely to choose the jar with the higher proportion of their preferred color.
02

Identify the sample statistic

The relevant sample statistic is the proportion of infants who chose the jar with a higher proportion of their preferred color (\( \hat{p} \)). \( \hat{p} = \frac{18}{24} = 0.75 \) .
03

Choose the correct test for p-value

A test using the normal distribution should be used to calculate the p-value. This is because we are dealing with a proportion in a population and have a sample size sufficient for the Central Limit Theorem to apply.
04

Compute the p-value

The p-value is calculated as follows: First, we calculate the test statistic, Z = \( \frac{ \hat{p} - p0}{\sqrt{\frac{p0(1 - p0)}{n}}} \), where \( p0 \) is the hypothesized proportion and n is the sample size. In this case, \( p0 \) would be 0.5 if infants choose randomly and n is 24. This gives us Z = \( \frac{0.75 - 0.5}{\sqrt{\frac{0.5(1 - 0.5)}{24}}} \) = 2.449 . The p-value is the probability that the Z-statistic is greater than 2.449, which can be found from the Z-table or a statistical calculator. For a one-sided test (as implied by the alternative hypothesis that infants are *more likely* to choose the higher proportion), the p-value = 0.0072.
05

Make a conclusion

We reject the null hypothesis if the p-value is less than the significance level \( \alpha \). Given \( \alpha = 0.05 \) and p-value = 0.0072, we reject the null hypothesis. Hence, there is enough evidence to suggest that infants are more likely to choose the jar with a higher proportion of their preferred color.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
The null hypothesis is a cornerstone of statistical testing. It represents the statement of no effect or no difference, which in the context of our baby probability study translates to the hypothesis that infants are not more likely to choose the jar with the higher proportion of their preferred color—in other words, that they would choose randomly. Formally stated, the null hypothesis \( H_0 \) in our study is that the true probability \( p \) of an infant selecting their preferred color is 0.5, regardless of the proportions presented in the jars.

Understanding the null hypothesis is crucial since it serves as the baseline assumption that will be tested against the evidence. If the observed data suggest that this baseline assumption is unlikely, then we might consider evidence in favor of the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis poses a contrasting statement to the null hypothesis and is what a researcher is typically trying to evidence. For the infants and probability experiment, the alternative hypothesis \( H_A \) is that infants are more likely to choose the jar with the higher proportion of their preferred color. It is the hypothesis that there is an actual effect or difference, and in statistical terms, it often represents the presence of a significant skew or deviation from what the null hypothesis proposes.

It is formulated based on the research question, and in our case, it reflects the idea that babies might exhibit probabilistic reasoning when selecting their favored color, essentially hypothesizing that infants have the ability to assess probabilities to some extent.
Sample Statistic
In statistical analysis, the sample statistic provides a snapshot of the data collected from the sample. It is a numerical measure that describes some aspect of the sample. When evaluating the behavior of infants in choosing lollipop colors, our sample statistic is the proportion \( \hat{p} \) of infants who selected the jar with a higher proportion of their preferred color. With 18 out of 24 infants picking the preferred color, our sample proportion \( \hat{p} \) is calculated as \( \hat{p} = \frac{18}{24} = 0.75 \).

This is a pivotal piece of our examination as it is the actual observable result that we'll compare against the hypothesized scenario provided by our null hypothesis.
P-value Calculation
Calculating the p-value is an essential step in hypothesis testing and helps us determine the significance of our sample statistic within the context of the null hypothesis. It represents the probability of observing our sample statistic, or something more extreme, assuming the null hypothesis is true. For our lollipop color choice experiment, we used a normal distribution to calculate the p-value since our sample size is large enough, and we are dealing with proportions that lend themselves to the Central Limit Theorem.

We then determined the test statistic (Z), which is a measure of how many standard deviations our sample statistic (\( \hat{p} \) is from the hypothesized proportion (\( p_0 \) under the null hypothesis. Subsequently, we can find the probability associated with this Z-value, which is our p-value, using statistical tables or software. The low p-value we calculated (\( 0.0072 \) in our study suggests that what we observed is quite unlikely under the null hypothesis, leading us to consider the alternative hypothesis.
Statistical Significance
Statistical significance is the determination of whether the observed difference or effect in our data is unlikely to have occurred just by chance, given the null hypothesis is true. It's an essential part of hypothesis testing because it helps us understand whether to reject the null hypothesis or not. We use a predetermined threshold called the significance level (commonly denoted as \( \alpha \) to make this determination. For the infant study, this level is set at \( \alpha = 0.05 \).

If our p-value is less than \( \alpha \) as it is in the lollipop color choice experiment (p-value = 0.0072), we have enough evidence to state that our findings are statistically significant. This means we reject the null hypothesis and accept the alternative hypothesis, inferring a high likelihood that infants can indeed grasp the concept of probability to favor the jar with a higher proportion of the color they prefer.

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Most popular questions from this chapter

In Exercises 6.28 to 6.31, use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of peanuts in mixed nuts, with \(n=100\) and \(\hat{p}=0.52\)

In Exercises 6.7 and 6.8 , compute the standard error for sample proportions from a population with the given proportion using three different sample sizes. What effect does increasing the sample size have on the standard error? Using this information about the effect on the standard error, discuss the effect of increasing the sample size on the accuracy of using a sample proportion to estimate a population proportion. A population with proportion \(p=0.75\) for sample sizes of \(n=40, n=300,\) and \(n=1000 .\)

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When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.04 .

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